1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating a real integral with a complex integral

  1. Dec 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following real integral using complex integrals:
    [itex]\int_0^\infty \frac{cos(2x)}{x^2+4}dx[/itex]

    2. Relevant equations
    Cauchy's Residue Theorem for simple pole at a: [itex]Res(f;a)=\displaystyle\lim_{z\rightarrow a} (z-a)f(z)[/itex]


    3. The attempt at a solution
    Since the function [itex]\frac{cos(2x)}{x^2+4}dx[/itex] is even, [itex]\int_0^\infty \frac{cos(2x)}{x^2+4}dx =\frac{1}{2} \int_{-\infty}^\infty \frac{cos(2x)}{x^2+4}dx[/itex]

    Consider the real part of [itex]f(z)=\frac{e^{2iz}}{z^2+4}[/itex]

    [itex]\int_0^\infty \frac{cos(2x)}{x^2+4}dx[/itex]=[itex]\frac{1}{2}Re(\displaystyle\lim_{R\rightarrow \infty}\int_{[-R,R]} \frac{e^{2iz}}{z^2+4})dz[/itex]

    We have [itex]\int_{[-R,R]}f(z)dz=\int_\gamma f(z)-\int_{\gamma R} f(z)dz[/itex]

    where [itex]\int_{\gamma R} f(z)dz[/itex] tends to zero as can be shown by the estimation theorem.

    To calculate [itex]\int_\gamma f(z)[/itex] we find the residue of f at the simple pole z=2i:

    [itex]Res(f;2i)=\lim_{z\rightarrow {2i}}(z-2i)\frac{e^{2iz}}{z^2+4}=\lim_{z\rightarrow {2i}}\frac{e^{2iz}+2ie^{2iz}(z-2i)}{2z}=\frac{1}{4ie^4}[/itex]

    This is obviously incorrect since there can be no imaginary part
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 7, 2013 #2

    pasmith

    User Avatar
    Homework Helper



    The integral is [itex]2 \pi i[/itex] times the residue.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculating a real integral with a complex integral
Loading...